> ketralnis is right that this only terminates if an algorithm exists, so the claim that this terminates is equivalent to the axiom of choice.
No my algorithm above will never terminate. There's a recursive call where the input never shrinks and it won't converge on a base case.
It's equivalent to saying
def f(x):
return f(x)
which is a pointless (but true) statement. The whole thing is distracting everyone.
Basically ketralnis is clarifying context which is sort of lost with my post. There's an axiom and do you believe in the axiom of not?
If the axiom can be proven then it is not an axiom. But a theorem. What I'm doing here is kind of pointless because we know it's an axiom by definition, the question is whether this axiom is valid or not. Attempting to prove the axiom is a a signal that I'm lacking clarity with the logic here.
No my algorithm above will never terminate. There's a recursive call where the input never shrinks and it won't converge on a base case.
It's equivalent to saying
which is a pointless (but true) statement. The whole thing is distracting everyone.Basically ketralnis is clarifying context which is sort of lost with my post. There's an axiom and do you believe in the axiom of not?
If the axiom can be proven then it is not an axiom. But a theorem. What I'm doing here is kind of pointless because we know it's an axiom by definition, the question is whether this axiom is valid or not. Attempting to prove the axiom is a a signal that I'm lacking clarity with the logic here.